InterviewSolution
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InterviewSolution
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What is the effect of the following ionisation process on the bond order of O2 ?O2 \(\rightarrow\) O2+ + e- |
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Answer» O2 → O2+ + e- O2 = (16e-) is KK (σ2s)2(σ*2s)2(σ2pz)2(π2px)2 = (π2py)2(π*2px)1(π*2py)1 Bond order = \(\frac{1}{2}\)(10 - 6) = \(\frac{4}{2}\) = 2 O2+ = (15e-) KK (σ2s)2( σ*2s)2( σ2pz)2(π2px)2 = (π2py)2(π*2px)1 Bond order = \(\frac{1}{2}\)(10 - 5) = \(\frac{5}{2}\) = 2.5 Thus, bond order of O2 which is 2, increases to 2.5 in \(O_2^+\). |
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